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There is a theorem stating that there is essentially one smooth structure on $R^n$ for every n other than 4. Does anybody know where i could find the proof of this? Not so much of what happens in dimension four, where there are infinitely many, but of the uniqueness in other dimensions? Thanks!

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  • $\begingroup$ See also the answers to mathoverflow.net/questions/24930/… $\endgroup$
    – j.c.
    May 17, 2010 at 17:01
  • $\begingroup$ For the high-dimensional case, a key part of the proof is the uniqueness of smooth structures on compact discs, which you get from the h-cobordism theorem. skupers's answer gets you there. $\endgroup$ Jan 14, 2015 at 4:33

3 Answers 3

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For $n \geq 5$ , this was first proven in Stallings' The piecewise-linear structure of Euclidean space. It actually proves the PL case and applies smoothing theory. Anyway, Theorem 5.1 of it says

Let $M^n$ be a contractible differentiable manifold which is 1-connected at infinity. If $n \geq 5$, then $M$ is diffeomorphic to Euclidean space $\mathbb{R}^n$.

A related result appears in Lashof's ICM address on smoothing theory. His Corollary says

Every contractible open topological manifold is smoothable.

and he follows by remarking that this smooth structure is unique if $n \geq 5$.

For $n = 2,3$, there is Moise, but his articles may be hard to read. There are now easier proofs using Kirby-Siebenmann techniques. For $n = 2$, one can use Hatcher's The Kirby torus trick for surfaces. For $n = 3$, one can use Hamilton's The triangulation of 3-manifolds in the PL case and apply smoothing theory.

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You can handle the case of $n \leq 3$ one at a time, and so the question really is about $n \geq 5$. Two important names in this regard are Kirby and Siebenmann. The Wikipedia article on the Hauptvermutung is a good place to start.

If M is an $n$-dimensional topological manifold (and $n \geq 5$), then $M$ admits a PL structure if and only if a special cohomology class, the Kirby-Siebenmann class, in $H^4(M; \mathbb{Z}_2)$ vanishes. If this class vanishes, then the different PL structures are parametrized up to concordance by $H^3(M; \mathbb{Z})$. (Note: The Wikipedia article on the Hauptvermutung assumes that $M$ is compact, but I don't believe that this is a necessary assumption.)

So what does this say about $M = \mathbb{R}^n$? Well, we already know that $\mathbb{R}^n$ has a PL structure, and since $H^3(\mathbb{R}^n; \mathbb{Z}_2)=0$, it follows that this structure is unique up to concordance. Since concordance implies diffeomorphism, and since every smooth structure gives us a PL structure, it follows that there can be only one smooth structure on $\mathbb{R}^n$ up to diffeomorphism.

Here are the main references (you can find them both here):

  1. Kirby and Siebenmann, On the triangulation of manifolds and the Hauptvermutung. Bull. Amer. Math. Soc. 75 1969 742--749.

  2. Kirby and Siebenmann, Foundational essays on topological manifolds, smoothings, and triangulations. Annals of Mathematics Studies 88 (1977). (I did some MathSciNet investigating, and the relevant essays are IV and V.)

This expository article by Rudyak, which I found through Wikipedia, also seems interesting.

Finally, I learned all of this from Scorpan's wonderful book, "The Wild World of 4-Manifolds".

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S. K. Donaldson and P. B. Kronheimer. The geometry of four- manifolds. Clarendon Press, New York, 1990.

Michael Freedman and Frank Quinn. Topology of 4-manifolds. Princeton University Press, Princeton, 1990.

These are recommended by Lee in his "Topology of Smooth Manifolds"

Edwin E. Moise. Geometric Topology in Dimensions 2 and 3. Springer-Verlag, New York, 1977.

James R. Munkres. Obstructions to the smoothing of piecewise differentiable homeomorphisms. Annals of Math., 72:521–554, 1960.

These are also recommended by Lee as proving the R^n (except for 4) claim.

I personally have not read them, but I, for now, trust Lee's judgement.

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  • $\begingroup$ Do any of these references discuss uniqueness of smooth structures on R^n for n>4? $\endgroup$
    – Tim Perutz
    Feb 22, 2010 at 15:55
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    $\begingroup$ This is not quite an answer to the question. Only the Moise and Munkres references are to answers. $\endgroup$ Jan 14, 2015 at 4:32

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